Looking for a counter example for non-connected intersection of descending chain of closed connected sets

Solution 1:

In $\mathbb{R}^{2}$ take two disjoint infinite lines $L_{1},L_{2}$ connected by vertical lines $B_{1},...$ (like a ladder). Then define $$A_{n}=L_{1}\cup L_{2}\cup \bigcup_{k\geq n}B_{k}.$$

These form a descending chain and their intersection will be $L_{1}\cup L_{2}$.

Solution 2:

The answer here can also be said without the quotient:

Let $X:=[-1,1]\cup\{0'\}$ where $0'$ is a new, fictive element, playing the role of a second origin, and define the topology on $[-1,1]$ as usual, and let $(-a,b)\setminus\{0\}\cup\{0'\}$ be a base for open neighborhoods of $0'$.

Now consider $Y_n:=[-1/n,1/n]\cup\{0'\}$.

Solution 3:

Try the following sets: $$Y_i = \bigg([-2,2]\times\mathbb{R}\bigg) - \bigg((-1,1)\times(-i,i)\bigg).$$

That is, $Y_i$ is a closed infinite strip with successively larger open boxes removed. The intersection $$\bigcap_{i=1}^\infty Y_i = ([-2,-1]\cup[1,2])\times\mathbb{R}$$ is disconnected, but each $Y_i$ is closed and connected.